is there a formula to find the total of all the didgets up to a #?

Question

for nstance 12=3 (1+2) the total up to 12 is 1+2+3+4+5+6+7+8+9+1+3
i need to find the total of all the didgets to 1,000,000,000

pleez help i need this for school, adn it starts in 50 min

i have to leave in 20 min

theres gotta be some kind of theory or equation that works for this, i just dont know what it is, pleeeeeeeez hurry

Answer 1

You have a billion numbers from 000,000,000 to 999,999,999. These are all the possible combinations of digits 0 to 9 at each position (units, tens, hundreds, thousands, tens of thousands etc... to hundreds of millions).

You have 100,000,000 numbers starting with a 0.
100,000,000 starting with a 1.
etc...
100,000,000 starting with a 9.

Since 0+1+2+3+4+5+6+7+8+9=45, this means the sum of all first digits is 45 x 100,000,000 = 4,500,000,000.

Similarly you will find that the sum of alll second digits (tens of millions) is also 4.5 billion. Same for third digits, etc... and for ninth digits(units).

So the sum of all digits from 0 to 999,999,999 is 9 times 45 billion = 40,500,000,000.

You need to add 1 for the number 1,000,000,000, so the answer is 40.5 billion + 1 = 40,500,000,001.

Answer 2

Look up triangular numbers. find the formular, digits for 1, for 2, ...

for n for n+1, check your definition you forgot 11 ==> 2

Total 11= 1+2+3+4+5+6+7+8+9+1+2
total 12= 1+2+3+4+5+6+7+8+9+1+2+3

T1=1
T2=1+2


Tn=x

Tn+1 =x + Sn+1

Answer 3

Yup!